Question

if you start with 120 - gram sample of radium. how much will be left after 44 days?

Explanation:

1. Explanation:

• First, we need to know the half - life of Radium. The half - life of Radium - 226 is approximately 1600 years. Since we are working with days, convert 1600 years to days. Assuming 365 days in a year, the half - life \(T = 1600\times365= 584000\) days.
• The radioactive decay formula is \(N = N_0(\frac{1}{2})^{\frac{t}{T}}\), where \(N_0\) is the initial amount, \(t\) is the time elapsed, and \(T\) is the half - life.
• Here, \(N_0 = 120\) grams, \(t = 44\) days, and \(T=584000\) days.
• Calculate the exponent \(\frac{t}{T}=\frac{44}{584000}\).
• Then find \((\frac{1}{2})^{\frac{44}{584000}}\).
• Finally, multiply by \(N_0\) to get the remaining amount \(N\).

Step 1: Calculate the exponent

\(\frac{t}{T}=\frac{44}{584000}\approx7.534\times10^{-5}\)

Step 2: Calculate the decay factor

\((\frac{1}{2})^{\frac{44}{584000}}=(\frac{1}{2})^{7.534\times 10^{-5}}\)
Using the formula \(a^x = e^{x\ln(a)}\), we have \((\frac{1}{2})^{7.534\times 10^{-5}}=e^{7.534\times 10^{-5}\ln(\frac{1}{2})}\)
\(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693\), so \(e^{7.534\times 10^{-5}\times(- 0.693)}=e^{-5.221\times10^{-5}}\approx0.9999478\)

Step 3: Calculate the remaining amount

\(N = N_0(\frac{1}{2})^{\frac{t}{T}}=120\times0.9999478\approx119.994\) grams

Answer:

Approximately \(119.994\) grams